Metamath Proof Explorer


Theorem ldualvsubcl

Description: Closure of vector subtraction in the dual of a vector space. (Contributed by NM, 27-Feb-2015)

Ref Expression
Hypotheses ldualvsubcl.f ⊢ F = LFnl ⁡ W
ldualvsubcl.d ⊢ D = LDual ⁡ W
ldualvsubcl.m ⊢ - ˙ = - D
ldualvsubcl.w ⊢ φ → W ∈ LMod
ldualvsubcl.g ⊢ φ → G ∈ F
ldualvsubcl.h ⊢ φ → H ∈ F
Assertion ldualvsubcl ⊢ φ → G - ˙ H ∈ F

Proof

Step Hyp Ref Expression
1 ldualvsubcl.f ⊢ F = LFnl ⁡ W
2 ldualvsubcl.d ⊢ D = LDual ⁡ W
3 ldualvsubcl.m ⊢ - ˙ = - D
4 ldualvsubcl.w ⊢ φ → W ∈ LMod
5 ldualvsubcl.g ⊢ φ → G ∈ F
6 ldualvsubcl.h ⊢ φ → H ∈ F
7 eqid ⊢ Scalar ⁡ W = Scalar ⁡ W
8 eqid ⊢ inv g ⁡ Scalar ⁡ W = inv g ⁡ Scalar ⁡ W
9 eqid ⊢ 1 Scalar ⁡ W = 1 Scalar ⁡ W
10 eqid ⊢ + D = + D
11 eqid ⊢ ⋅ D = ⋅ D
12 7 8 9 1 2 10 11 3 4 5 6 ldualvsub ⊢ φ → G - ˙ H = G + D inv g ⁡ Scalar ⁡ W ⁡ 1 Scalar ⁡ W ⋅ D H
13 eqid ⊢ Base Scalar ⁡ W = Base Scalar ⁡ W
14 7 lmodring ⊢ W ∈ LMod → Scalar ⁡ W ∈ Ring
15 4 14 syl ⊢ φ → Scalar ⁡ W ∈ Ring
16 ringgrp ⊢ Scalar ⁡ W ∈ Ring → Scalar ⁡ W ∈ Grp
17 15 16 syl ⊢ φ → Scalar ⁡ W ∈ Grp
18 13 9 ringidcl ⊢ Scalar ⁡ W ∈ Ring → 1 Scalar ⁡ W ∈ Base Scalar ⁡ W
19 15 18 syl ⊢ φ → 1 Scalar ⁡ W ∈ Base Scalar ⁡ W
20 13 8 grpinvcl ⊢ Scalar ⁡ W ∈ Grp ∧ 1 Scalar ⁡ W ∈ Base Scalar ⁡ W → inv g ⁡ Scalar ⁡ W ⁡ 1 Scalar ⁡ W ∈ Base Scalar ⁡ W
21 17 19 20 syl2anc ⊢ φ → inv g ⁡ Scalar ⁡ W ⁡ 1 Scalar ⁡ W ∈ Base Scalar ⁡ W
22 1 7 13 2 11 4 21 6 ldualvscl ⊢ φ → inv g ⁡ Scalar ⁡ W ⁡ 1 Scalar ⁡ W ⋅ D H ∈ F
23 1 2 10 4 5 22 ldualvaddcl ⊢ φ → G + D inv g ⁡ Scalar ⁡ W ⁡ 1 Scalar ⁡ W ⋅ D H ∈ F
24 12 23 eqeltrd ⊢ φ → G - ˙ H ∈ F